In a given trapezium $ ABCD $ with $ AB$ parallel to $ CD $ and $ AB > CD $, the line $ BD $ bisects the angle $ \angle ADC $. The line through $ C $ parallel to $ AD $ meets the segments $ BD $ and $ AB $ in $ E $ and $ F $, respectively. Let $ O $ be the circumcenter of the triangle $ BEF $. Suppose that $ \angle ACO = 60^{\circ} $. Prove the equality \[ CF = AF + FO .\]
Problem
Source: Middle European Mathematical Olympiad 2012 - Individuals I-3
Tags: geometry, trapezoid, circumcircle, geometry proposed
ACCCGS8
15.09.2012 10:51
$\angle EDC = \angle ADE = \angle DEC = \alpha \implies EC = DC \implies EC = FA$. Also $O$ is the circumcentre of $BEF$ so $EO = FO$. Standard angle chasing gives $\angle AFC = 2\alpha, \angle EFO = 90 - \alpha, \angle FEB = \alpha, \angle BEO = 90 - 2\alpha \implies \angle AFO + \angle FEO = 180 \implies \angle AFO = \angle CEO$ So $AFO$ and $CEO$ are congruent by SAS $\implies \angle FCO = \angle FAO \implies AFOC$ cyclic. So $\angle AFO = 120 \implies 2\alpha + 90 - \alpha = 120 \implies \alpha = 30$. Thus $EFO$ is equilateral, so $CF = CE + EF = DC + FO = AF + FO$, as required.
reveryu
03.02.2017 10:49
@ACCCGS8 some correction, BEO=2(alpha)-90
Kimchiks926
25.03.2022 01:52
Trivial angle chasing reveals that $AD=AB$ and $FE=FB$, whereas trivial length chasing gives us that $AF=CE$.
The main observation is that $\triangle AFO = \triangle CEO$, which follows from the angle chasing giving that $\angle CEO = \angle AFO$. This implies several things. First of all, quadrilateral $ACOF$ is cyclic, since $\angle OAF = OCF$. Second of all, $\triangle ACO$ is equilateral, since $CO=AO$ combined with $\angle ACO=60^{\circ}$. Now obviously $\triangle OEF$ is equilateral, meaning that $EF =FO$ and the problem is solved.